How do you find the vertices, asymptote, foci and graph #y^2/25-x^2/16=16#?

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Answer 1 · 2017-02-22T05:23:29

Vertices: #(0, +-20)#
Asymptotes: #y = +- 5/4 x#
Foci: #(0, +-4sqrt(41))#

First put the hyperbola in standard form when the largest denominator is under the #y^2#: #(y-k)^2/a^2 - (x-h)^2/b^2 = 1#

Multiple all terms by #1/16# to get a #1# on the right-side:
#y^2/(25*16) - x^2/(16*16) = 1#; #y^2/(400) - x^2/16^2 = 1#; #y^2/20^2 - x^2/16^2 = 1#

Center: #(h,k) = (0,0)#

#a = 20, b = 16#

#c^2 = a^2+b^2 = 656, c = sqrt(656) = sqrt(16*41) = 4sqrt(41)#

Vertices: #(h, k+-a) = (0, +-20)#

Asymptotes: #y-k = +-a/b(x-h); y = +-20/16x; y = +-5/4x#

Foci: #(h,k+-c) = (0, +-4sqrt(41))#

To graph:

Create a box centered around the center that is length #a# vertically above and below the center and length #b# horizontally above and below the center (use dashed lines).
Sketch the asymptote lines (use a dashed line).
Place dots at the vertices and foci.
Sketch a curve from each vertex, staying inside of the asymptote lines.
Graphing Hyperbolas